The function $f$ has the property that for each real number $x$ in its domain, $1/x$ is also in its domain and \[
f(x) + f\left(\frac{1}{x}\right) = x.
\]What is the largest set of real numbers that can be in the domain of $f$?

(a) ${\{x\mid x\ne0\}}$

(b) ${\{x\mid x<0\}}$

(c) ${\{x\mid x>0\}}$

(d) ${\{x\mid x\ne-1\ \text{and}\ x\ne0\ \text{and}\ x\ne1\}}$

(e) ${\{-1,1\}}$
Answer: The conditions on $f$ imply that both \[
x = f(x) + f\displaystyle\left(\frac{1}{x}\displaystyle\right)\]and \[\frac{1}{x} = f\left(\frac{1}{x}\right) +
f\displaystyle\left(\frac{1}{1/x}\displaystyle\right) = f\displaystyle\left(\frac{1}{x}\displaystyle\right) + f(x).
\]Thus if $x$ is in the domain of $f$, then $x = 1/x$, so $x = \pm 1$.

The conditions are satisfied if and only if $f(1)=1/2$ and $f(-1)=-1/2$.  Hence the answer is $\boxed{E}$.